## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$f(x)=(x+1)(x+3)(x-2)$ Zeros: $-3,\ \ -1,\ \ 2,$ and all with multiplicity $1$.
Let us consider that $m$ is a factor of the constant term and $n$ is a factor of the leading coefficient. Then the potential zeros can be expressed by the possible combinations as: $\dfrac{m}{n}$. We see from the given polynomial function that it has at most $3$ real zeros as the degree is $3$. The possible factors $m$ of the constant term and $n$ of the leading coefficient are: $m=\pm 1, \pm 2, \pm 3, \pm 6$ and $n=\pm 1$ Therefore, the possible rational roots of $f(x)$ are: $\dfrac{m}{n}=\pm 1, \pm 2, \pm 3, \pm 6$ We test with synthetic division. We will try $x+1$. $\left.\begin{array}{l} -1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr} 1 &2 &-5&-6\\\hline &-1&-1 &6\\\hline 1&1 &-6& |\ \ 0\end{array}$ Thus, we have: $f(x)=(x+1)(x^{2}+x-6) \\=(x+1)(x+3)(x-2)$ Zeros: $-3,\ \ -1,\ \ 2,$ and all with multiplicity $1$.